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EXPLANATION OF PRASEODYMIUM IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) June 15 , 2015 Praseodymium is a chemical element with symbol Pr and atomic number 59. However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this image of Praseodymium including the following ground state electron configuration: 1s22s22p63s23p63d104s24p64d105s25px25py25pz24f36s2 According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (eV) of praseodymium (from (E1 to E5 ) are the following: E1 = 5.47 , E2 = 10.55, E3 = 21.62, E4 = 38.98, and E5 = 57.53 . For understanding better such ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my paper of 2008. EXPLANATION OF E1 = 5.47 eV = -E(6s2) + E(6s1) Here the E(6s2) represents the binding energy of 6s2, while the E(6s1) represents the binding energy of 6s1. The charges (-57e) of the 57 electrons of (1s22s22p63s23p63d104s24p64d105s25p64f3) screen the nuclear charge (+59e) and for a perfect screening we would have ζ = 2. However the electrons of 6s penetrate the 4f and 5p and lead to the deformation of electron clouds. Thus ζ > 2 . Note that the 6s2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(6s2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 6s1 consists of one electron, we apply the Bohr formula to write E(6s1) = (-13.6057)ζ2/n2 Therefore E1 = 5.47 eV = -E(6s2) + E(6s1) = - (16.95)ζ + 4.1) / n2 Then using n = 6 the above equation can be written as (13.6057)ζ2 - (16.95)ζ - 192.82 = 0 Then solving for ζ we get ζ = 4.44 > 2 . Of course the two electrons of opposite spin (6s2) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy, which seems to be like a simple electric repulsion of the Coulomb law. This situation of a vibration energy due to an electromagnetic interaction indeed occurs, because the peripheral velocity of a spinning electron is faster than the speed of light, which invalidates Einstein’s theory of special relativity. (See my FASTER THAN LIGHT). However under the influence of invalid relativity and in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that it is due to the Coulomb repulsion. Under such fallacious ideas I published my paper of 2008 EXPLANATION OF E2 = 10.55 eV = -E(6s1) As in the case of E1 the charges (-57e) of 57 electrons of the following electron configuration (1s22s22p63s23p63d104s24p64d105s25px25py25pz24f3) screen the nuclear charge (+59e) and for a perfect screening we would have the same ζ = 2. However the one electron of 6s1 penetrates the 4f3 and the 5p6 and leads to the deformation of electron clouds. Thus we would have ζ > 2. Here the E(6s1) represents the binding energy of (6s1) given by applying the Bohr formula as E2 = 10.55 eV = -Ε(6s1) = - ( -13.6057)ζ2 /62 Then solving for ζ we get ζ = 5.28 > 2 . Here the 5.28 > 2 means that after the ionization the one electron of 6s breaks more the symmetry and leads to a great deformation of electron clouds. EXPLANATION OF E3 = 21.62 eV = -E(4f1) Here the E(4f1) represents the binding energy of the first electron (4f 1) given by applying the Bohr formula as E3 = 21.62 = -E(4f1) = -(-13.6057)ζ2 /n2 The charges (-54e) of the 54 electrons of the following electron configuration (1s22s22p63s23p63d104s24p64d105s25px25py25pz2) screen the nuclear charge (+59e) and for a perfect screening we would have an effective ζ = 5. However in the absence of the two outer electrons of 6s2, the electrons of 4f repel the electrons of 5p and lead to the deformation of electron clouds . Thus ζ > 5. Therefore using n = 4 we get ζ = 5.04 > 5 . Here we observe a strange phenomenon, because the electrons of 4f seem to be in inner orbital than that of the 5p electrons. In fact, the electrons of 4f under n = 4 repel the electrons of 5p in such a way that the deformed cloud of 4f electrons are always at the opposite position with respect to the deformed electron cloud of 5p. EXPLANATION OF E4 = 38.98 eV = -E(4f1) Here the E(4f1) represents the binding energy of the second electron (4f1) given by applying the Bohr formula as E4 = 38.98 = -E(4f1) = -(-13.6077)ζ2 /n2 As in the case of E3 the charges (-54e) of the following electron configuration (1s22s22p63s23p63d104s24p64d105s25px25py2 5pz2) screen the nuclear charge (+59e) and for a perfect screening we would have the same effective ζ = 5. However after the ionizations the second electron of 4f1 brakes more the symmetry and leads to a greater deformation of electron clouds. Thus ζ > 5. Then using n = 4 we get ζ = 6.77 > 5. ' ' EXPLANATION OF E5 = 57.53 eV = - E(4f1) Here the E(4f1) represents the binding energy of the third electron (4f1) given by applying the Bohr formula as E5 = 57.53 = -E(4f1) = -(-13.6057)ζ2 /n2 As in the case of E4 the charges (-54e) of the following electron configuration (1s22s22p63s23p63d104s24p64d105s25px25py25pz2) screen the nuclear charge (+59e) and for a perfect screening we would have the same effective ζ = 5. However after the ionizations the third electron of 4f1 brakes more the symmetry and leads to a greater deformation of electron clouds. Thus ζ >> 5. Then using n = 4 we get ζ = 8.23 >> 5. Category:Fundamental physics concepts